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SashulF [63]
3 years ago
6

A mile is 5280 feet. between which two integers is the elevation in miles

Mathematics
2 answers:
Cloud [144]3 years ago
7 0
<span>-4 and-5 hope it helps </span>
vampirchik [111]3 years ago
7 0
That would be 2 I think 
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Algebra 2A Help !<br><br> Tadakimasu ~
Ludmilka [50]
Function is p(x)=(x-4)^5(x^2-16)(x^2-5x+4)(x^3-64)

first factor into (x-r1)(x-r2)... form

p(x)=(x-4)^5(x-4)(x+4)(x-4)(x-1)(x-4)(x^2+4x+16)
group the like ones
p(x)=(x-4)^8(x+4)^1(x-1)^1(x^2+4x+16)

multiplicity is how many times the root repeats in the function
for a root r₁, the root r₁ multiplicity 1 would be (x-r₁)^1, multility 2 would be (x-r₁)^2 

so

p(x)=(x-4)^8(x+4)^1(x-1)^1(x^2+4x+16)
(x-4)^8 is the root 4, it has multiplicity 8
(x-(-4))^1 is the root -4 and has multiplicity 1
(x-1)^1 is the  root 1 and has multiplity 1
(x^2+4x+16) is not on the real plane, but the roots are -2+2i√3 and -2-2i√3, each multiplicity 1 (but don't count them because they aren't real

baseically

(x-4)^8 is the root 4, it has multiplicity 8
(x-(-4))^1 is the root -4 and has multiplicity 1
(x-1)^1 is the  root 1 and has multiplity 1

7 0
3 years ago
Which function f (x) , graphed below, or g (x) , whose equation is g (x) = 3 cos 1/4 (x + x/3) + 2, has the largest maximum and
Leya [2.2K]

Answer:

g(x), and the maximum is 5

Step-by-step explanation:

for given function f(x), the maximum can be seen from the shown graph i.e. 2

But for the function g(x), maximum needs to be calculated.

Given function :

g (x) = 3 cos 1/4 (x + x/3) + 2

let x=0 (as cosine is a periodic function and has maximum value of 1 at 0 angle)

g(x)= 3 cos1/4(0 + 0) +2

    = 3cos0 +2

     = 3(1) +2

     = 3 +2

     = 5 !

7 0
2 years ago
Jeremy traveled 345 miles to visit his cousin in north carolina. if he traveled at a rate of 60 miles per hour, how long did the
Orlov [11]
The trip took 5 hours and 45 mins
3 0
3 years ago
Help me to answer this question pl​s
attashe74 [19]

Problem 1

Draw a straight line and plot P anywhere on it. Use the compass to trace out a faint circle of radius 8 cm with center P. This circle crosses the previous line at point Q.

Repeat these steps to set up another circle centered at Q and keep the radius the same. The two circles cross at two locations. Let's mark one of those locations point X. From here, we could connect points X, P, Q to form an equilateral triangle. However, we only want the 60 degree angle from it.

With P as the center, draw another circle with radius 7.5 cm. This circle will cross the ray PX at location R.

Refer to the diagram below.

=====================================================

Problem 2

I'm not sure why your teacher wants you to use a compass and straightedge to construct an 80 degree angle. Such a task is not possible. The proof is lengthy but look up the term "constructible angles" and you'll find that only angles of the form 3n are possible to make with compass/straight edge.

In other words, you can only do multiples of 3. Unfortunately 80 is not a multiple of 3. I used GeoGebra to create the image below, as well as problem 1.

8 0
2 years ago
James and Sarah went out to lunch. The price of lunch for both of them was $60. They tipped their server 10% of that amount.
kvasek [131]

Answer:

$66

Step-by-step explanation:

$60 *0.1 is 6. $60+6 is 66. That is the price of the whole thing

Hope this helps plz mark brainliest :D

3 0
2 years ago
Read 2 more answers
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